Question: Find the largest constant $m,$ so that for any positive real numbers $a,$ $b,$ $c,$ and $d,$
\[\sqrt{\frac{a}{b + c + d}} + \sqrt{\frac{b}{a + c + d}} + \sqrt{\frac{c}{a + b + d}} + \sqrt{\frac{d}{a + b + c}} > m.\]
Explanation: By GM-HM applied to 1 and $\frac{a}{b + c + d},$
\[\sqrt{1 \cdot \frac{a}{b + c + d}} \ge \frac{2}{\frac{1}{1} + \frac{b + c + d}{a}} = \frac{2a}{a + b + c + d}.\]Similarly,
\begin{align*}
\sqrt{\frac{b}{a + c + d}} &\ge \frac{2b}{a + b + c + d}, \\
\sqrt{\frac{c}{a + b + d}} &\ge \frac{2c}{a + b + c + d}, \\
\sqrt{\frac{d}{a + b + c}} &\ge \frac{2d}{a + b + c + d}.
\end{align*}Adding up all these inequalities, we get
\[\sqrt{\frac{a}{b + c + d}} + \sqrt{\frac{b}{a + c + d}} + \sqrt{\frac{c}{a + b + d}} + \sqrt{\frac{d}{a + b + c}} \ge \frac{2a + 2b + 2c + 2d}{a + b + c + d} = 2.\]The only we can get equality is if
\begin{align*}
a &= b + c + d, \\
b &= a + c + d, \\
c &= a + b + d, \\
d &= a + b + c.
\end{align*}Adding these equations, we get $a + b + c + d = 3(a + b + c + d),$ so $a + b + c + d = 0,$ which is impossible.  Thus, equality is not possible.

However, by setting $a = c = 1$ and $b = d = \epsilon,$ where $\epsilon$ is a small positive number, then
\[\sqrt{\frac{a}{b + c + d}} + \sqrt{\frac{b}{a + c + d}} + \sqrt{\frac{c}{a + b + d}} + \sqrt{\frac{d}{a + b + c}} = 2 \sqrt{\frac{1}{1 + 2 \epsilon}} + 2 \sqrt{\frac{\epsilon}{2 + \epsilon}}.\]As $\epsilon$ approaches 0, the expression approaches 2.  Thus, we can make the expression arbitrarily close to 2, so $m = \boxed{2}.$